operational amplifiers chapter 2 - mcgill universitygrober4/pdfs/ecse330/2. op amp...op amp...
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© 2009 G. W. Roberts Op Amp Circuits, slide 1
304-330 Introduction To Electronic Circuits
Operational Amplifiers Chapter 2
© 2009 G. W. Roberts Op Amp Circuits, slide 2
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 3
304-330 Introduction To Electronic Circuits
The Universal Analog Building Block: Main Terminals
The op amp shown connected to dc power supplies (with noise decoupling capacitors).
€
C
€
C
(C)
The Op Amp Symbol
Three terminals shown plus power supply connections. Op amps have other terminals that is used to tuned
its non-ideal behavior.
© 2009 G. W. Roberts Op Amp Circuits, slide 4
304-330 Introduction To Electronic Circuits
Electrical Equivalent Op Amp Circuit
• Input/Output Impedance:
=
€
Zin− =
v1
i1= ∞ Zin
+ =v2
i2= ∞ Zout =
voio v1 −v2 =K
io = ix
= 0
Output impedance is derived by setting the input sources to some constant value, and a current into the output port and deriving the corresponding output voltage.
vout= f v1,v2( )
© 2009 G. W. Roberts Op Amp Circuits, slide 5
304-330 Introduction To Electronic Circuits
Electrical Equivalent Op Amp Circuit
• One output, two inputs:
– If circuit is linear, then
=
€
vout = f v1,v2( )
€
vout = A2v2 + A1v1
€
A1v1 + A2v2
© 2009 G. W. Roberts Op Amp Circuits, slide 6
304-330 Introduction To Electronic Circuits
Equivalent Signal Representation:CM and DM Signals
• Any two node voltages (relative to analog ground) can be expressed as a linear combination of their corresponding DM & CM signal components.
€
v1 = vI ,cm −vI ,d2
v2 = vI ,cm +vI ,d2
v1
v2
Node voltages:
Solving For CM & DM signals:
€
vI ,cm =v1 + v22
vI ,d = v2 − v1
© 2009 G. W. Roberts Op Amp Circuits, slide 7
304-330 Introduction To Electronic Circuits
Creating The Differential Source Arrangement
• Op Amp circuits must be driven with the source arrangement shown here.
€
A1v1 + A2v2
© 2009 G. W. Roberts Op Amp Circuits, slide 8
304-330 Introduction To Electronic Circuits
Creating The Differential Source Arrangement
• By design, A1= -A2 = A:
vout =A2v2 +A1v1 v1=vI ,cm−vI ,d 2v2=vI ,cm+vI ,d 2
= A1+A2( )vI ,cm +12A1−A2( )vI ,d
€
vout = 0 ⋅vI ,cm + A1 ⋅vI ,d = A v2 − v1( )
Amplification depends only on the DM signal and not the CM signal
€
A1v1 + A2v2
© 2009 G. W. Roberts Op Amp Circuits, slide 9
304-330 Introduction To Electronic Circuits
Creating The Differential Source Arrangement
€
vout = 0 ⋅vI ,cm + A2 ⋅vI ,d = A2 v2 − v1( )
Amplification depends only on the DM and not the CM signal: A 2-input system becomes a 1-input system.
€
A1v1 + A2v2
€
vout = A1v1 + A2v2= A1 vI ,cm − vI ,d 2( ) + A2 vI ,cm + vI ,d 2( )
= A1 + A2( )vI ,cm +12A2 − A1( )vI ,d
By design, A1= -A2 :
© 2009 G. W. Roberts Op Amp Circuits, slide 10
304-330 Introduction To Electronic Circuits
Op Amp Internal Architecture• Example: Calculate the output
voltage as a function of the input v1 and v2 using the following parameters:
• Solution: €
Gm = 10mAV, R = 10 kΩ, µ = 100
€
A1 =vov1 v2 =0
= −µGmR = −104 VV
A2 =vov2 v1 =0
= µGmR = 104 VV
€
∴vO = 104 v2 − v1( ) VV
© 2009 G. W. Roberts Op Amp Circuits, slide 11
304-330 Introduction To Electronic Circuits
A1v1 + A2v2
Common-Mode Rejection Ratio CMRR Measure Of Gain Balance
• In practice, achieving balanced gains A1=-A2 is difficult.
• Define the gains
• A m e a s u r e o f t h e g a i n imbalance can be described as
Design Goal:
CMRR ≡ AdAcm
We want large CMRR (Ad>>Acm) for 2 input system to become a 1
input system.
€
vout = A1 + A2( )vI ,cm +12A2 − A1( )vI ,d
€
vout = AcmvI ,cm + AdvI ,d
Acm = A1 + A2 Ad =A2 − A1
2
where
© 2009 G. W. Roberts Op Amp Circuits, slide 12
304-330 Introduction To Electronic Circuits
CMRR Example• The gain from the + and –
input terminals of the op amp was found to be
• What is the CM and DM gains? Write an equation describing the input-output behavior and compute the CMRR.
Solution:
A1v1 + A2v2
€
A1 = −10,000 VV
A2 = +10,500 VV
€
Acm = A1 + A2 = −10,000 +10,500 = 500 VV
Ad =12A2 − A1( ) =
12
10,500 +10,000( ) = 10,250 VV
€
vout = −10,000 ⋅v1 +10,500 ⋅v2 or
vout = 500 ⋅vI ,cm +10,250 ⋅vI ,d
CMRR =10,250
500= 20.5
© 2009 G. W. Roberts Op Amp Circuits, slide 13
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 14
304-330 Introduction To Electronic Circuits
The Inverting Configuration• Op amps are not used on their own; rather, they
are used in some form of feedback arrangement.
R1 & R2 close feedback around op amp
© 2009 G. W. Roberts Op Amp Circuits, slide 15
304-330 Introduction To Electronic Circuits
Analysis Of The Inverting Configuration
• According to op amp behavior:
• By KCL at the negative input terminal of op amp we can write:
• Feedback structure:
€
vo = −Av1
€
vI − v1R1
+vO − v1R2
= 0
⇒ v1 =R2
R1 + R2vI +
R1R1 + R2
vO
€
R1R1 + R2
€
−A+
€
R2R1 + R2
€
vI
€
vOv1
© 2009 G. W. Roberts Op Amp Circuits, slide 16
304-330 Introduction To Electronic Circuits
Analysis Of The Inverting Configuration
• Solv ing fo r input -output transfer function:
• In the limit, as A -> ∞, we write
• Also note that in the limit we have a virtual short to ground,
€
G =vOvI
=−R2 R1
1+ 1+ R2 R1( ) A
€
G =vOvI
= −R2R1
€
v2 − v1 =vOA A→∞
= 0
© 2009 G. W. Roberts Op Amp Circuits, slide 17
304-330 Introduction To Electronic Circuits
Quick Ideal Analysis Of Op Amp Circuits
• Using the virtual short concept, we can rapidly derive the input-output transfer function of op amp circuits.– C i r c l e d n u m b e r s
i n d i c a t e a n a l y s i s sequence.
• Note that the virtual short concept arrives because of the negative feedback principle; it is not a function of the op amp itself.
© 2009 G. W. Roberts Op Amp Circuits, slide 18
304-330 Introduction To Electronic Circuits
Accounting For Finite Open Loop Gain
• Direct circuit analysis reveals
• Solving, we write
€
G =vOvI
=−R2 R1
1+ 1+ R2 R1( ) A
€
v1 = −vOA
(1)
i1 =vI − −vO A( )
R1(2)
i2 = i1 (3)
vO = −vOA−R2R1
vI +vOA
# $ %
& ' (
(4)• As long as
the open loop gain has little effect on the closed loop gain and we get
€
1+ R2 R1( ) A << 1
€
G ≈ −R2 R1
© 2009 G. W. Roberts Op Amp Circuits, slide 19
304-330 Introduction To Electronic Circuits
Why Construct Amplifiers Using Amplifiers?
• Op amps (an amplifier) are used to construct other amplifiers as this arrangement provides a level of manufacturing robustness that is needed for mass production.– This is an example of
negative feedback.Amplifier (op amp)
Amplifier
© 2009 G. W. Roberts Op Amp Circuits, slide 20
304-330 Introduction To Electronic Circuits
Example: Open Loop Gain Dependence
• Let us investigate the relative CL gain error for the inverting amplifier as a function of op amp gain:
€
ε =Gactual − Gideal
Gideal
×100%
A |Gideal| |Gactual| ε v-
103 100 90.83 -9.17% -9.08 mV104 100 99.00 -1.00% -0.99 mV105 100 99.90 -0.10% -0.10 mV∞ 100 100 0% 0
€
R1 = 1 kΩ
€
R1 = 100 kΩ
€
v −
© 2009 G. W. Roberts Op Amp Circuits, slide 21
304-330 Introduction To Electronic Circuits
Example: Open Loop Gain Dependence
This example illustrates how robust the closed loop amplifier gain is to changes in the op amp gain.
A |Gactual|A 105 99.90
A-A/2 105-105/2 99.80Δ -105/2 -0.1% -50% -0.1%
€
R1 = 1 kΩ
€
R1 = 100 kΩ
€
v −
A -50% change in op amp gain lead to relatively small change -0.1% in the closed loop gain.
© 2009 G. W. Roberts Op Amp Circuits, slide 22
304-330 Introduction To Electronic Circuits
Input / Output Resistance
€
Rin =vIi1
=vI
vI R1= R1
€
Rout = vO ix= 0
With vI=0, v2-v1=0, vo= A(v2-v1)=0 then
€
Rout = 0ix
= 0
© 2009 G. W. Roberts Op Amp Circuits, slide 23
304-330 Introduction To Electronic Circuits
Example: Inverting AmplifierCalculate the currents in the op amp circuit shown to the right assuming the circuit is driven by a 1 V source.Solution:
€
v1 = 0 V
i1 = 1 V1 kΩ = 1 mA
i2 = i1 = 1 mAvO = −i2 ×10 kΩ = −10 V
iL = −10 V1 kΩ = −10 mA
iO = iL − i2 = −10 mA−1 mA = −11 mA
vovI
=−10 V
1 V= −10 V
V iL
i1=−10 mA
1 mA= −10 A
A PL
PI=−10 V( ) −10 mA( )
1 V( ) 1 mA( )= 100 mW
mW
Note the op amp output current; comes from supplies.
© 2009 G. W. Roberts Op Amp Circuits, slide 24
304-330 Introduction To Electronic Circuits
Design An Amplifier For A Microphone Application
• Provide a simple electrical model of a microphone assuming the source impedance is 1 MΩ:
© 2009 G. W. Roberts Op Amp Circuits, slide 25
304-330 Introduction To Electronic Circuits
Design An Amplifier For A Microphone Application
• Select the values of the feedback resistors so that the magnitude of the gain is 100 V/V:
© 2009 G. W. Roberts Op Amp Circuits, slide 26
304-330 Introduction To Electronic Circuits
Large Resistor Values• Resistor values larger than 10
MΩ are impractical.• Must select resistors values from
a standard set for cost reasons:– see EIA (Electronic Industries
Associat ion) STANDARD RESISTOR VALUES listings.
• Resistors are not made exact, they are made within a tolerance (0.1%, 1%, 5%, 10% and 20%)– The smaller the tolerance, the
greater the cost.
© 2009 G. W. Roberts Op Amp Circuits, slide 27
304-330 Introduction To Electronic Circuits
Creating Resistors With Precise Values
• When precise resistor values are required, we connect in series or parallel, standard low cost components.
series/parallel combination
© 2009 G. W. Roberts Op Amp Circuits, slide 28
304-330 Introduction To Electronic Circuits
Can You Design an Amplifier With A Gain Magnitude of 100 V/V Using
Small Resistor Values? • Student Solution:
• What is the cost of this design versus the previous design with a large resistor value?− Assume each op amp cost $5 and resistors cost $0.5 each.
© 2009 G. W. Roberts Op Amp Circuits, slide 29
304-330 Introduction To Electronic Circuits
High Input Impedance With Small Feedback Resistors
• In some sensor applications, amplification is required but the amplifier must have a large input resistance (say 1 MΩ).
• This, in turn, causes the feedback resistors to be impractical or unrealizable if the gain is 10 or greater, i.e.,
• Need a new design.
€
Rin = R1
€
G = −R2R1
€
R2 = G × Rin
© 2009 G. W. Roberts Op Amp Circuits, slide 30
304-330 Introduction To Electronic Circuits
High Input Impedance With Small Feedback Resistors
• As the input resistance is established by R1, the design choices for the feedback resistance becomes
€
Rin = R1
€
G = −R2R1
1+R4R2
+R4R3
#
$ %
&
' (
€
R2 1+R4R2
+R4R3
"
# $
%
& ' = G × Rin
© 2009 G. W. Roberts Op Amp Circuits, slide 31
304-330 Introduction To Electronic Circuits
High Input Impedance With Small Feedback Resistors
• Reducing the number of degrees of freedom, select:
• This results in the following for the remaining resistor:
• Here we see R3 is less than that required by the input resistance condition.
€
R4 = R2 = R1 = Rin
€
R3 =Rin
G − 2
© 2009 G. W. Roberts Op Amp Circuits, slide 32
304-330 Introduction To Electronic Circuits
Current Amplifier WithFloating Load
• A current amplifier pushes a current through the load resistor independent of the load resistance.
Load is floating: No terminals
connected to ground.
€
Rin = 0Rout = ∞
€
H =i4iI
= 1+R2R3
"
# $
%
& '
© 2009 G. W. Roberts Op Amp Circuits, slide 33
304-330 Introduction To Electronic Circuits
A Digital Calculator
© 2009 G. W. Roberts Op Amp Circuits, slide 34
304-330 Introduction To Electronic Circuits
An Analog Calculator
© 2009 G. W. Roberts Op Amp Circuits, slide 35
304-330 Introduction To Electronic Circuits
The Weighted Summer Circuit
€
vo = −Rf
R1v1 +
Rf
R2v2 + +
Rf
Rn
vn#
$ %
&
' (
• A weighted summer is made possible by the virtual ground at the input of the op amp. – Multiple currents are combined.
© 2009 G. W. Roberts Op Amp Circuits, slide 36
304-330 Introduction To Electronic Circuits
A Dual-Sign Weighted Summer
• Cascading two multiple-inverter amplifiers enables the sign of the signal weight to be positive or negative depending on where the input is applied.€
vo =Ra
R1
Rc
Rb
v1 +Ra
R2
Rc
Rb
v2 −Rc
R3v3 −
Rc
R4v4
© 2009 G. W. Roberts Op Amp Circuits, slide 37
304-330 Introduction To Electronic Circuits
Assignment #2 (partial) • Sedra/Smith:
– 2.5, 2.6, 2.8, 2.9, 2.12, 2.15, 2.20, 2.34, 2.41
© 2009 G. W. Roberts Op Amp Circuits, slide 38
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 39
304-330 Introduction To Electronic Circuits
The Noninverting ConfigurationInput is applied to + op amp terminal:
Circuit Analysis
€
vO = 1+R2R1
"
# $
%
& ' vI
© 2009 G. W. Roberts Op Amp Circuits, slide 40
304-330 Introduction To Electronic Circuits
The Noninverting ConfigurationKCL at – op amp terminals:
€
R1R1 + R2
€
−A+
€
1
€
vI
€
vO
€
−v1R1
+vO − v1R2
= 0⇒ v1 =R1
R1 + R2vO
€
v1
op amp behavior:
€
vO = −A v1 − vI( )
Feedback structure:
© 2009 G. W. Roberts Op Amp Circuits, slide 41
304-330 Introduction To Electronic Circuits
The Noninverting ConfigurationInput / Output Impedance
Input Impedance:
Output Impedance:
€
Rin = ∞
€
Rout = 0
€
Rin
€
Rout
© 2009 G. W. Roberts Op Amp Circuits, slide 42
304-330 Introduction To Electronic Circuits
Example: Noninverting AmplifierCalculate the currents in the op amp circuit shown to the right assuming the circuit is driven by a 1 V source.Solution:
€
v1 = vI = 1 ViI = 0 A
i1 = 1 V1 kΩ = 1 mA
i2 = i1 = 1 mAvO = 1 V + i2 × 9 kΩ = 10 V
iL = 10 V1 kΩ = 10 mA
iO = iL + i2 = 10 mA +1 mA =11 mA
vovI
=10 V1 V
= 10 VV
iLiI
=10 mA
0 A= ∞
AA
PLPI
=10 V( ) 10 mA( )
1 V( ) 0 A( )= ∞
mWmW
© 2009 G. W. Roberts Op Amp Circuits, slide 43
304-330 Introduction To Electronic Circuits
The Voltage Follower
• The voltage follower circuit is a special case of the noninverting configuration; all of the output signal is feedback to the op amp input.
• As the name implies, the voltage gain is unity.
© 2009 G. W. Roberts Op Amp Circuits, slide 44
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 45
304-330 Introduction To Electronic Circuits
Electronics To Aid Patient Care
© 2009 G. W. Roberts Op Amp Circuits, slide 46
304-330 Introduction To Electronic Circuits
Body Sensors For Medical Monitoring
© 2009 G. W. Roberts Op Amp Circuits, slide 47
304-330 Introduction To Electronic Circuits
ECG Heart Monitoring Transducers
© 2009 G. W. Roberts Op Amp Circuits, slide 48
304-330 Introduction To Electronic Circuits
Common-Mode Level Problem
• The human body can charge up to static voltage levels as high as 15,000 volts by simply walking across a carpeted floor, or 5,000 volts by walking across a linoleum floor.
• The potential difference between a charged human body and an object retaining an insignificant charge can range from a few hundred volts to as high as 30,000 volts.
• To amplify body signals, one cannot use a straightforward amplifier approach.
~
± 10,000 V
10 mV
vIN
t 0
10,000 V
10 mV vo
t 0 VCC
VDD VDD
VDD << 10,000 V
© 2009 G. W. Roberts Op Amp Circuits, slide 49
304-330 Introduction To Electronic Circuits
Difference Amplifier
Difference amplifier is a combination of inverting and noninverting amplifier.
Inverting amplifier
Noninverting amplifier
Difference amplifier
€
vO1 = −R2R1vI1
€
vO2 = 1+R2R1
"
# $
%
& ' vI 2
vo =α vI 2 − vII( )
© 2009 G. W. Roberts Op Amp Circuits, slide 50
304-330 Introduction To Electronic Circuits
Analyzing The Difference Amplifier
• Set vI2=0 and solve for output in terms of input vI1.
• Set vI1=0 and solve for output in terms of input vI2.
€
vO1 = −R2R1vI1
€
vO2 = 1+R2R1
"
# $
%
& '
R4R3 + R4
"
# $
%
& ' vI 2
• Using superposition, we can write:
€
vO = vO1 + vO2 = −R2R1vI1 + 1+
R2R1
#
$ %
&
' (
R4R3 + R4
#
$ %
&
' ( vI 2
© 2009 G. W. Roberts Op Amp Circuits, slide 51
304-330 Introduction To Electronic Circuits
Difference Amplifier With DM & CM Signals
€
vO = −R2R1vI1 + 1+
R2R1
#
$ %
&
' (
R4R3 + R4
#
$ %
&
' ( vI 2
€
vI1 = vI ,cm −vI ,d
2 and vI 2 = vI ,cm +
vI ,d
2
€
⇒ vO = −R2R1
vI ,cm −vI ,d2
$ % &
' ( ) + 1+
R2R1
$
% &
'
( )
R4R3 + R4
$
% &
'
( ) vI ,cm +
vI ,d2
$ % &
' ( )
∴vO = −R2R1
+ 1+R2R1
$
% &
'
( )
R4R3 + R4
$
% &
'
( )
+
, -
.
/ 0 vI ,cm +
12R2R1
+ 1+R2R1
$
% &
'
( )
R4R3 + R4
$
% &
'
( )
+
, -
.
/ 0 vI ,d
© 2009 G. W. Roberts Op Amp Circuits, slide 52
304-330 Introduction To Electronic Circuits
Goal Of A Difference Amplifier
• The goal of a difference amplifier is to have Acm = 0 while producing a non-zero Ad.
• This means the first term above must equate to 0, i.e.
€
vO = −R2R1
+ 1+R2R1
#
$ %
&
' (
R4R3 + R4
#
$ %
&
' (
)
* +
,
- . vI ,cm +
12R2R1
+ 1+R2R1
#
$ %
&
' (
R4R3 + R4
#
$ %
&
' (
)
* +
,
- . vI ,d
€
−R2R1
+ 1+R2R1
#
$ %
&
' (
R4R3 + R4
#
$ %
&
' ( = 0
⇒R2R1
= 1+R2R1
#
$ %
&
' (
R4R3 + R4
#
$ %
&
' (
∴R1R2
=R3R4
€
vO = 0 ⋅vI ,cm +R2R1vI ,d
• Substituting back into the above equation we get:
© 2009 G. W. Roberts Op Amp Circuits, slide 53
304-330 Introduction To Electronic Circuits
Input Resistance Of Difference Amplifier
• With R3=R1 and R4=R2, the input resistance is
• The feedback resistor can be found form the gain
• As before, a large voltage gain and a large input resistance signifies a large feedback resistor, which in turn limits its practicality.
• Once again, we need a new circuit where the DM gain and input resistance is decoupled (Instrumentation Amplifier).
€
Rid = 2R1
€
R2 =12Rid ×G
© 2009 G. W. Roberts Op Amp Circuits, slide 54
304-330 Introduction To Electronic Circuits
Difference Amplifier With BufferBuffer (Av=1)
Difference amplifier has unity gain and provide difference
function only.
Adding buffers to front-end increases input resistance of diff. amp.
© 2009 G. W. Roberts Op Amp Circuits, slide 55
304-330 Introduction To Electronic Circuits
Instrumentation Amplifier First Step In Evolution
Noninverting amplifier provides high-impedance input and additional gain.
CM signal is also amplified making the diff amp work harder.
Difference amplifier has unity gain and provide difference
function only.
© 2009 G. W. Roberts Op Amp Circuits, slide 56
304-330 Introduction To Electronic Circuits
Instrumentation Amplifier Second Step In Evolution
• Floating the shared resistor R1 eliminates any matching concerns associated with the front-end amplifiers.
Removing the GND connection
eliminates CM signal
component.
© 2009 G. W. Roberts Op Amp Circuits, slide 57
304-330 Introduction To Electronic Circuits
Instrumentation Amplifier Ideal Op Amp Analysis
© 2009 G. W. Roberts Op Amp Circuits, slide 58
304-330 Introduction To Electronic Circuits
Instrumentation Amplifier Design Goal
• Typically, instrumentation amplifier is designed with all the gain in the front-end stage and the difference amplifier having a gain of unity.
€
R
€
R
€
R
€
R€
R
€
R
€
vO = 1+RR1
"
# $
%
& ' vId
=
€
vId = vI 2 − vI1
-
+ +
-
© 2009 G. W. Roberts Op Amp Circuits, slide 59
304-330 Introduction To Electronic Circuits
Assignment #2 (partial) • Sedra/Smith:
– 2.5, 2.6, 2.8, 2.9, 2.12, 2.15, 2.20, 2.34, 2.41– 2.44, 2.50, 2.51, 2.53, 2.60, 2.72, 2.74
© 2009 G. W. Roberts Op Amp Circuits, slide 60
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 61
304-330 Introduction To Electronic Circuits
The Inverting Configuration With General Impedances
• Replacing R1 and R2 in the inverting amplifier with general impedance allows for the development of several other interesting amplifier structures.
€
Vo
VI
s( ) = −Z2 s( )Z1 s( )
© 2009 G. W. Roberts Op Amp Circuits, slide 62
304-330 Introduction To Electronic Circuits
Example
€
Vo
VI
s( ) = −Z2 s( )Z1 s( )
• Consider the general form of the TF of the inverter:
• If
• then
Z1 s( )=R1 Z2 s( )=1
1R2+sC2
VoVI
s( )=−Z2 s( )Z1 s( )
=−
11R2+sC2
R1
=−R2 R1
1+sC2R2
⇒K =−R2 R1 ω3−dB =1
C2R2
© 2009 G. W. Roberts Op Amp Circuits, slide 63
304-330 Introduction To Electronic Circuits
The Inverting Integrator Time-Domain Description
• Replacing Z1 and Z2 by R and 1/sC, respectively, leads to a very important circuit called the Miller Integrator.
• This circuit has the ability to integrate an input signal.
Add vO(0-) term if nonzero.
© 2009 G. W. Roberts Op Amp Circuits, slide 64
304-330 Introduction To Electronic Circuits
The Inverting Integrator Frequency Domain Description
• For physical frequencies, replace s by jω and write
• The magnitude and phase can then be written as
or, in dB we write
€
Vo
VI
s( ) = −1sCR
→s= jω
Vo
VI
jω( ) = −1
jωCR
€
Vo
VI
jω( ) =1
ωCR ∠Vo
VI
jω( ) = +90
€
dB = 20log10Vo
VI
jω( ) = 20log101
ωCR
dB = 20log101CR
− 20log10ω
ω t =1CR
© 2009 G. W. Roberts Op Amp Circuits, slide 65
304-330 Introduction To Electronic Circuits
The Inverting Integrator Frequency Domain Description
• We also see that a pole exists at DC, i.e.,
• What does this mean in terms of the circuit?– Feedback path opens on
account of the capacitor acts as an open circuit.
– Op amp is operated in open loop, where the gain is assume infinite (ideal).
• Circuit is not usable in its present form! must be fixed.
€
Vo
VI
s( ) = −1sCR
→s=0
Vo
VI
s( ) = ∞
- +
R
C
€
vO± vI
- +
R
€
vO± vI
DC Equivalent Model
Miller Integrator
Any input will experience infinite gain.
∞
© 2009 G. W. Roberts Op Amp Circuits, slide 66
304-330 Introduction To Electronic Circuits
Differential Equation Solver:The Analog Computer
• Op amps circuits (even before the invention of transistors) were used to solve differential equations.
© 2009 G. W. Roberts Op Amp Circuits, slide 67
304-330 Introduction To Electronic Circuits
Differential Equation Solver:The Analog Computer
• Can you see the application of the integrator in this circuit?
- +
C1
R1 - +
R5
R5 - +
C2
R4
R3 R2
€
x t( )
€
y t( )
€
αd2y t( )dt 2
+ βdy t( )dt
+ γy t( ) + ρdx t( )dt
= 0
© 2009 G. W. Roberts Op Amp Circuits, slide 68
304-330 Introduction To Electronic Circuits
The Miller Integrator With Damping• Connecting a feedback resistor in parallel with
C can help to eliminate the effects of a pole at DC and the instabilities associated with it.
• Consider the circuit transfer function:
• A pole is now located at
• A small value of RF places the pole at a large frequency, reducing the range of integrator operation; however, a large RF introduces a larger offset due to op amp offset and bias currents.– Trade off between AC and DC operation.
€
Vo
VI
s( ) = −RF R
1+ sCRF
€
s = −1
CRF
ω0 1
€
RF R
€
Vo
VI
€
ω p =1
CRF€
ω t =1CR€
DC ∝ RF
© 2009 G. W. Roberts Op Amp Circuits, slide 69
304-330 Introduction To Electronic Circuits
Compare Ideal Vs Damped Integrator Operation
• Physical frequencies larger than ωp will be correctly integrated by the damped integrator.
• Frequencies less than ωp will be amplified, not integrated.
ω0
€
RF R
€
Vo
VI
€
ω p =1
CRF
ideal Int.
damped Int.
ω t =1CR
© 2009 G. W. Roberts Op Amp Circuits, slide 70
304-330 Introduction To Electronic Circuits
Integration Compare In Time Domain
Ideal Integrator
Damped Integrator
Step Input
© 2009 G. W. Roberts Op Amp Circuits, slide 71
304-330 Introduction To Electronic Circuits
Integration Compare In Time Domain
• For error less than 1%, a damped integrator will correctly integrate a signal over a time span between 0 and CRF/10 seconds.
• How does this result compare with the result we found in the frequency domain?
t vo
ideal Int.
damped Int.
τ I ,max =CRF10
about 1% error
© 2009 G. W. Roberts Op Amp Circuits, slide 72
304-330 Introduction To Electronic Circuits
The Op Amp Differentiator
• Rearranging the position of the R and C changes the circuit to a differentiator.
© 2009 G. W. Roberts Op Amp Circuits, slide 73
304-330 Introduction To Electronic Circuits
The Op Amp Differentiator Frequency Domain Description
• For physical frequencies, replace s by jω and write
• The magnitude and phase can then be written as
or, in dB we write
€
Vo
VI
s( ) = −sCR →s= jω
Vo
VI
jω( ) = − jωCR
€
Vo
VI
jω( ) =ωCR ∠Vo
VI
jω( ) = −90
€
dB = 20log10Vo
VI
jω( ) = 20log10ωCR
dB = 20log10CR + 20log10ω
ω t =1CR
© 2009 G. W. Roberts Op Amp Circuits, slide 74
304-330 Introduction To Electronic Circuits
Assignment #2 (partial) • Sedra/Smith:
– 2.5, 2.6, 2.8, 2.9, 2.12, 2.15, 2.20, 2.34, 2.41– 2.44, 2.50, 2.51, 2.53, 2.60, 2.72, 2.74 – 2.79, 2.80, 2.85, 2.89
© 2009 G. W. Roberts Op Amp Circuits, slide 75
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 76
304-330 Introduction To Electronic Circuits
The Many Layers Of Analog DesignAmplifier Topology, Diff Gain & CM Gain
Slew-Rate, large-signal bandwidth
DC Offsets
3-dB Bandwidth (small-signal)
• High-performance analog design is the result of correcting many small errors along the way.
Input Bias Currents
Large-signal opertion
© 2009 G. W. Roberts Op Amp Circuits, slide 77
304-330 Introduction To Electronic Circuits
Voltage Transfer Characteristic
+ - +
-
€
vO = VO,OFF
VDD
VSS
+ -
±
+-
VDD
VSS
€
vO = 0
−VOS
vId
VSS
VDD
vO
0,0
vI,max vI,min vId
VSS
VDD vO
0,0
vI,max vI,min
Ideal Nonideal
-VOS
Offset Present Removing Offset With Series V
Note sign change!
VO,OFF A
A
€
−VOS =VO,OFFA
© 2009 G. W. Roberts Op Amp Circuits, slide 78
304-330 Introduction To Electronic Circuits
Some Facts About Op Amp Offset Voltage
• Over mass production, the offset is not fixed or constant, rather will vary over a range of values.– It is a random variable and is not known before
manufacture.• Offset is an important measurement of the repeatability of a
manufacturing process.
vId
VSS
VDD vO
0,0 measure
© 2009 G. W. Roberts Op Amp Circuits, slide 79
304-330 Introduction To Electronic Circuits
Op Amp Voltage Offset Model
• The offset associated with an op amp is modeled with a voltage source VOS placed in series with the +ve terminal of the op-amp.
• An external voltage source of –VOS can be used to cancel the effect of the offset.
+ -
±
VDD
VSS
€
vO = 0
VOS±
−VOS Actual Op Amp
Offset-Free Op Amp
+ -
vId
VSS
VDD vO
0,0 -VOS
VO,OFF
A
© 2009 G. W. Roberts Op Amp Circuits, slide 80
304-330 Introduction To Electronic Circuits
Impact Of Offsets On CL Circuit
• The effect of offsets on the closed-loop circuit behavior can be derived using circuit analysis.– Set the input equal to zero and calculate output behavior.
• Here we see the offset is amplifier by the gain of the circuit; a 5 mV offset can be increase by a factor as large as 1000, resulting in an output of +5V.
vI
VSS
VDD
vO
0,0 -VOS
VO,OFF
CL Slope=(1+R2/R1)
OL Slope=A
© 2009 G. W. Roberts Op Amp Circuits, slide 81
304-330 Introduction To Electronic Circuits
Impact Of Offsets On CL Circuit
• Input signal conditions are set by the system design and cannot be changed.
• Offsets reduce the dynamic range of a circuit.– In order to obtain a distortion less
sinusoidal signal (symmetrical swing), the input signal amplitude must be reduced to compensate for the shift in the output DC level.
vI
VSS
VDD
vO
0,0
VO,OFF
vO
t
t
vI
© 2009 G. W. Roberts Op Amp Circuits, slide 82
304-330 Introduction To Electronic Circuits
Offset Trimming• Offsets can be removed by
trimming each and every op amp circuit.
• One technique is to use the offset-nulling terminals of an op amp.
• Typically, a potentiometer is connected across the offset-nulling terminals and tuned so that the output offset voltage is eliminated (set to zero).
© 2009 G. W. Roberts Op Amp Circuits, slide 83
304-330 Introduction To Electronic Circuits
Capacitively Coupled Circuit
• Another approach in which to minimize the DC offset effect on a closed-loop circuit is to connect a capacitor in series with the input terminals.
• At DC, the op amp circuit behaves as a unity gain follower, hence the offset does not experience any signal gain at the output.
• The drawback of this approach is the noninverting amplifier takes on a high-pass response with the 3-dB break point located at 1/CR1.
– Input signals must therefore be above this break frequency to prevent any unexpected attenuation.
VoVI
=sCR21+ sCR1
© 2009 G. W. Roberts Op Amp Circuits, slide 84
304-330 Introduction To Electronic Circuits
Input Bias Currents• Due to the electronic nature
of an op amp, two separate currents are pulled (or pushed) into the op amp input terminals.– These currents are
essential for the op amp operation; they cannot b e b l o c k e d b y a capacitor.
• This is modeled by two current sources IB1 and IB2.
© 2009 G. W. Roberts Op Amp Circuits, slide 85
304-330 Introduction To Electronic Circuits
Input Bias Currents• Op amp data sheets usually
specify the average value of these two bias currents as well as there expected difference:
Input bias current:
Input offset current:
€
IB =IB1 + IB 22
€
IOS = IB1 − IB 2
© 2009 G. W. Roberts Op Amp Circuits, slide 86
304-330 Introduction To Electronic Circuits
Calculating Bias Current Impact On Closed-Loop Operation
• Analysis of the CL behavior reveals that the bias current creates a DC offset at the output given by
€
VO = IB1R2
© 2009 G. W. Roberts Op Amp Circuits, slide 87
304-330 Introduction To Electronic Circuits
Series Resistor Compensation
• The effect of the DC bias current can be reduced by the addition of a series compensation resistor R3. – R3 has no effect on the signal transfer function but lowers
the virtual ground level by -IB2xR3.
€
VO = −IB 2R3 + R2 IB1 − IB 2R3R1
#
$ %
&
' (
Add compensation
resistor
© 2009 G. W. Roberts Op Amp Circuits, slide 88
304-330 Introduction To Electronic Circuits
Series Resistor Compensation
• Next, let us consider IB1 and IB2 are equal to the average bias current IB, we write
• The offset voltage can be made to go to zero if and only if
€
VO = −IB 2R3 + R2 IB1 − IB 2R3R1
#
$ %
&
' (
IB1 = IB 2 = IB
⇒VO = IB −R3 + R2 −R2R3R1
#
$ %
&
' (
€
R3 =R1R2R1 + R2
• The compensation resistor R3 should be set to the parallel equivalent of the DC resistance connect to the inverting input terminal.
© 2009 G. W. Roberts Op Amp Circuits, slide 89
304-330 Introduction To Electronic Circuits
Series Resistor Compensation• Now, let us consider the the
impact of the di f ference between IB1 and IB2 are on the offset voltage, we write
• Now with the previous result,
• We obtain the offset voltage as
VO = −IB2R3 + R2 IB1 − IB2R3R1
⎛⎝⎜
⎞⎠⎟
IB1 = IB + I OS 2IB2 = IB − I OS 2
⇒VO = − IB − I OS 2( )R3 + IB + I OS 2( )R2 − IB − I OS 2( ) R2R3R1
€
R3 =R1R2R1 + R2
• As IOS is generally quite small, we see that the resultant offset due to bias currents can be kept quite low.
€
VO = IOSR2
© 2009 G. W. Roberts Op Amp Circuits, slide 90
304-330 Introduction To Electronic Circuits
Bias Current Compensation Example
• Decoupling a bias path eliminates the dependency of the offset on the resistance in that signal path.
• Recall that a DC bias path is essential for op amp operation - be careful with C.
C blocks IB trough R1 C blocks IB trough R1
© 2009 G. W. Roberts Op Amp Circuits, slide 91
304-330 Introduction To Electronic Circuits
Effects Of Op Amp Offset Voltage On Miller Integrator Operation
• An op amp offset excites the pole at DC and causes the output to build towards infinity.
• Output will therefore saturate at VDD.
© 2009 G. W. Roberts Op Amp Circuits, slide 92
304-330 Introduction To Electronic Circuits
Effects Of Input Bias and Offset Currents On Miller Integrator Operation
• Input bias current also excite the pole at DC causing the output to ramp towards VDD.
© 2009 G. W. Roberts Op Amp Circuits, slide 93
304-330 Introduction To Electronic Circuits
Assignment #2 (partial) • Sedra/Smith:
– 2.5, 2.6, 2.8, 2.9, 2.12, 2.15, 2.20, 2.34, 2.41,– 2.44, 2.50, 2.51, 2.53, 2.60, 2.72, 2.74, – 2.79, 2.80, 2.85, 2.89,– 2.98, 2.104, 2.106,
© 2009 G. W. Roberts Op Amp Circuits, slide 94
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 95
304-330 Introduction To Electronic Circuits
Transient / Frequency Response
+ - ± +
-
Step-Response
+ - ± +
-
Sinusoidal-Response
€
vO
€
ωo
Decay rate related to op amp parameters
€
vO
€
ω
€
ω3dB
€
A jω( )
3-dB BW related to op amp parameters
€
vO t( )
€
t
© 2009 G. W. Roberts Op Amp Circuits, slide 96
304-330 Introduction To Electronic Circuits
Response To A Sinusoidal Input
• Linear system behavior can be measured through the application of a sinusoidal signal.– If linear, only the amplitude and phase of the
sinusoidal signal will change as it passes through a system.
– If nonlinear, the shape of the sinusoidal will change.» We’ll return to this idea a bit later.
G(s) +
- ~ vi(t) vo(t)
Ain sin ωot +ϕ in( )
Aout sin ωot +ϕout( )A1 sin ωot +ϕout( )+ A2 sin 2ωot +ϕout( )+…
Aout sin ωot +ϕout( )
© 2009 G. W. Roberts Op Amp Circuits, slide 97
304-330 Introduction To Electronic Circuits
Frequency Response
• As the input frequency changes, so too does the amplitude of the output signal.
• The amplitude and phase of the output signal are related according to:
Ain = 1.0 V Fin = 1 kHZ Ain = 1.0 V Fin = 2 kHZ Ain = 1.0 V Fin = 3 kHZ Ain = 1.0 V Fin = 4 kHZ Ain = 1.0 V Fin = 5 kHZ
Aout = 1.0 V Fout = 1 kHZ Aout = 1.0 V Fout = 2 kHZ Aout = 0.7 V Fout = 3 kHZ Aout = 0.6 V Fout = 4 kHZ Aout = 0.5 V Fout = 5 kHZ
Aout = G s( ) s= jω Ain = G( jω) Ain
ϕout =ϕin +∠G s( ) s= jω =ϕin +∠G jω( )
G(s) +
- ~ vi(t) vo(t)
© 2009 G. W. Roberts Op Amp Circuits, slide 98
304-330 Introduction To Electronic Circuits
System Step Response
• Electronic circuits are also designed with a desired step response in mind.
G(s) +
- ± u(t) vo(t)
vo(t) = L−1 L u t( )"# $%×L g t( )"# $%{ }= L−1 1s ×G s( )
'()
*+,
© 2009 G. W. Roberts Op Amp Circuits, slide 99
304-330 Introduction To Electronic Circuits
Frequency Description Of Op Amp Response
• Due to op amp finite gain and bandwidth, the basic op amp circuits shown previously deviate from their expected or ideal behavior.
• Depending on region of operation (i.e. frequencies) we can use different models of operation.
A(s) Vo Vi
+
-
ω0 ωb
1
AO
ωt
€
A s( ) =Vo
Vi
=AO
1+ sωb
€
A jω( ) = A s( )s= jω
=AO
1+ jωωb
A jω( ) ≈ AOωb
ω≡ωt
ω
€
A jω( ) = 1@ω ≈ω t
© 2009 G. W. Roberts Op Amp Circuits, slide 100
304-330 Introduction To Electronic Circuits
Piecewise Approximate Op Amp Frequency Response Behavior
A(s) Vo Vi
+
- ω
0 ωb 1
AO
ωt
A jω( ) =AO ω ≤ωb
AOωb
ωω >ωb
⎧
⎨⎪
⎩⎪
=AO ω ≤ωb
ω t
ωω >ωb
⎧
⎨⎪
⎩⎪
A jω( ) ≈ AOωb
ω≡ωt
ω
€
A jω( ) = 1@ω ≈ω t
A jω( ) ≈ AO for ω <ωb
© 2009 G. W. Roberts Op Amp Circuits, slide 101
304-330 Introduction To Electronic Circuits
Op Amp Equivalent CircuitOne Pole Model
A(s) Vo Vi
+
- = Vi
Vo
R
C AOVi
+
-
+
- ±
ω0
1
AO
€
Vo
Vi
€
ωb =1RC
€
ω t =AORC
€
A s( ) =Vo
Vi
=AO
1+ sωb
=AO
1+ sRC
A jω( ) ≈ AOωRC
for ω >>ωb
A jω( ) ≈ AO for ω <ωb
± v1
+
- v1
© 2009 G. W. Roberts Op Amp Circuits, slide 102
304-330 Introduction To Electronic Circuits
Frequency Response Of The 741 General Purpose Internally Compensated Op Amp
Op amps single-pole response have very little
3-dB bandwidth!
3-dB BW Unity-gain frequency
DC Gain
€
A s( ) =105
1+ s2π ×10
© 2009 G. W. Roberts Op Amp Circuits, slide 103
304-330 Introduction To Electronic Circuits
Frequency Response Of A Closed-Loop Amplifier
• The frequency response behavior of an op amp will influence the circuit in which it is configured.– The overall closed-loop effect will be different
depending on circuit configuration.
Inverting amplifier Noninverting amplifier €
vO1 = −R2R1vI1
€
vO2 = 1+R2R1
"
# $
%
& ' vI 2
© 2009 G. W. Roberts Op Amp Circuits, slide 104
304-330 Introduction To Electronic Circuits
Frequency Response Of A Inverting Amplifier
Recall from an earlier analysis,
Substituting
We will get
Now, for AO >> 1+ R2/R1, we write
€
VOVI
s( ) =−R2 R1
1+ 1+ R2 R1( ) A s( )
A s( )= AO1+s ωb
;ωt =AOωb
VOVI
s( )= −R2 R1
1+1+R2 R1( )
AO+s 1+R2 R1( )
AOωb
VOVI
s( )= −R2 R1
1+s 1+R2 R1( )
ωt
When compared to single-pole model,
We get the model parameters:€
Af s( ) =ADC
1+ s ω3−dB
€
ADC = −R2 R1
ω3−dB =ω t1+ R2 R1( )
ω0 1
ADC
w3-dB
© 2009 G. W. Roberts Op Amp Circuits, slide 105
304-330 Introduction To Electronic Circuits
Frequency Response Of A Noninverting Amplifier
Recall from an earlier analysis,
Substituting
We will get
Now, for AO >> 1+ R2/R1, we write
€
VOVI
s( ) =1+ R2 R1( )
1+ 1+ R2 R1( ) A s( )
VOVI
s( )=1+R2 R1( )
1+1+R2 R1( )
AO+s 1+R2 R1( )
AOωb
VOVI
s( )=1+R2 R1( )
1+s 1+R2 R1( )
ωt
When compared to single-pole model,
We get the model parameters:€
Af s( ) =ADC
1+ s ω3−dB
€
ADC =1+ R2 R1
ω3−dB =ω t1+ R2 R1( )
ω0 1
ADC
w3-dB
A s( )= AO1+s ωb
;ωt =AOωb
© 2009 G. W. Roberts Op Amp Circuits, slide 106
304-330 Introduction To Electronic Circuits
Frequency response of an amplifier with a nominal gain of +10 V/V.
Frequency Response Behavior Of Two Amplifiers In Closed Loop
Noninverting amplifier Inverting amplifier
€
ADC =1+ R2 R1
ω3−dB =ω t1+ R2 R1( )
€
ADC = −R2 R1
ω3−dB =ω t1+ R2 R1( )
© 2009 G. W. Roberts Op Amp Circuits, slide 107
304-330 Introduction To Electronic Circuits
Gain-BW Trade-off Rule
Unity-gain frequency
DC Gain 1
3-dB BW 1
DC Gain 1 x 3-dB BW 1 DC Gain 2 x 3-dB BW 2
DC Gain 2
3-dB BW 2
© 2009 G. W. Roberts Op Amp Circuits, slide 108
304-330 Introduction To Electronic Circuits
Closed-Loop Gain - BW Trade-Off(linear scale)
BW0 0 ω t
Gain ⋅BW =ω t
ω t10
Gain
110
100
ω t100
© 2009 G. W. Roberts Op Amp Circuits, slide 109
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 110
304-330 Introduction To Electronic Circuits
Op Amp Output DC Limits
• Op amps, like all electronic devices operate over a range of input signals, then saturates or ceases to operate in the intended fashion.– The output has both voltage and current limits.– The user must stay below these limits if the expected
behavior is to be obtained.
DC Sweep Input DC Transfer Characteristic
+ - ± +
-
€
vO
VDD
VSS
vIN vIN vO,MIN
vO,MAX vO
0,0 vIN,max vIN,min
vO
IMIN
IMAX iO
0,0
vO,max
€
iOvO,min
© 2009 G. W. Roberts Op Amp Circuits, slide 111
304-330 Introduction To Electronic Circuits
Op Amp Input/Output DC Limits
• As a simplified rule of thumb, the output voltage will saturate when the input exceeds the range given by
vIN ,min,vIN ,max{ } < VSSAo,VDDAo
⎧⎨⎩
⎫⎬⎭
vIN
VSS
VDD
vO
0,0 vIN,max
vIN,min Slope=Ao
DC Transfer Characteristic
© 2009 G. W. Roberts Op Amp Circuits, slide 112
304-330 Introduction To Electronic Circuits
741 Op Amp Input/Output Limits
• The 741 op amp has a nominal DC gain of 105 V/V.• For ±15 V supplies, the input limits are:
vI
-15 V
15 V
vO
0,0 vIN,max vIN,min
vIN ,min,vIN ,max{ } < −15105 , 15
105⎧⎨⎩
⎫⎬⎭= −150 µV,+150 µV{ }
Slope=Ao=105
© 2009 G. W. Roberts Op Amp Circuits, slide 113
304-330 Introduction To Electronic Circuits
Modeling Op-amp Large-Signal Operation
• When op amps are used in a feedback configuration, the output op amp limits translate to the input terminals of the circuit.
• Consider the -10 V/V inverting amplifier:– Here the output is shown to saturate at ±13 V; – The amp. input limits are therefore ±13 V / 10 = ±1.3 V.
+13V
−13V
+1
© 2009 G. W. Roberts Op Amp Circuits, slide 114
304-330 Introduction To Electronic Circuits
Hard-Limit Versus Soft-Limit
• Op-amps generally have a soft-limiting behavior instead of a hard limit.
• The soft limiting behavior causes a light amount of distortion at the output of an amplifier.– More realistic.
vO,min
vO,max
vIN,max
vIN,min
vIN
vO
Hard Limit
vIN
vO
Soft Limit
vO,max
vIN,min
vO,min
vIN,max
© 2009 G. W. Roberts Op Amp Circuits, slide 115
304-330 Introduction To Electronic Circuits
Step Response Of Linear 1st-Order CL System
• The step response of a linear system follows a predictable behavior set by the system poles and zeros.– Doubling the input level, doubles the output
response.
vO,1(t) = 1− e−ω tt( ) ⋅V
vO,2 (t) = 1− e−ω tt( ) ⋅2V
V
0
2V
A(s) ≈ ω t
s
© 2009 G. W. Roberts Op Amp Circuits, slide 116
304-330 Introduction To Electronic Circuits
Step Response Of Actual 1st-Order CL System
• When the input exceeds the system input limit, the output no longer follows the expected linear behavior.
• Instead, one will observe that the output is constrained by
– This maximum limit is called the amplifier slew-rate (SR).
dvO t( )dt
≤ SR
vO,1(t) = 1− e−ω tt( ) ⋅V
vO,2 (t) = 1− e−ω tt( ) ⋅2V
V
0
2V
vO t( ) ≈ SR ⋅ t
A(s) ≈ ω t
s
© 2009 G. W. Roberts Op Amp Circuits, slide 117
304-330 Introduction To Electronic Circuits
What Causes Slew-Rate Limiting?
• If the input differential level exceeds some level, say VIN,max, then the following amplifier stage see a constant input level of VIN,max, regardless of other changes at the input.
• The input maximum can be expressed in terms of two op amp parameters, ωt and SR, i.e.,
V
0
2V
VIN ,max =SRω t
vO,1(t) = 1− e−2π fbwt( ) ⋅V
vO t( ) ≈ SR ⋅ tVIN ,max
A(s) ≈ ω t
s
VIN ,max
−VIN ,max+
- +1
© 2009 G. W. Roberts Op Amp Circuits, slide 118
304-330 Introduction To Electronic Circuits
Large-Signal Sinusoidal Response
• If then
Linear response
SR Limited response
1 fM
dvO t( )dt max
≤ SR ⇒ ddt
V̂ sin 2π fMt( )⎡⎣ ⎤⎦max
≤ SR
vIN t( ) = V̂ sin 2π fMt( ) vO t( ) =vIN t( ) , dvo t( )
dt max
≤ SR
triangle vIN t( )⎡⎣ ⎤⎦ , dvo t( )dt max
> SR
⎧
⎨
⎪⎪
⎩
⎪⎪
vO t( ) = V̂ sin 2π fMt( )
∴2π fM ⋅V̂ ≤ SR
© 2009 G. W. Roberts Op Amp Circuits, slide 119
304-330 Introduction To Electronic Circuits
Large-Signal Sinusoidal Response
• For a sine wave input, we see the conditions for linear operation must satisfy the following
– The larger the amplitude, the lower the frequency that can be applied to the amplifier input.
• For a fixed amplitude, fM is called the full-power bandwidth.
Linear response
SR Limited response
1 fM
vIN t( ) = V̂ sin 2π fMt( ) vO t( ) =vIN t( ) 2π fM ⋅V̂ ≤ SR
triangle vIN t( )⎡⎣ ⎤⎦ 2π fM ⋅V̂ > SR
⎧⎨⎪
⎩⎪
2π fM ⋅V̂ ≤ SR
© 2009 G. W. Roberts Op Amp Circuits, slide 120
304-330 Introduction To Electronic Circuits
Output Voltage Level – Gain and BW Trade-Off
BW0 0 ω tω t10
Gai
n
1
10
100
ω t100
Gain ⋅BW =ω t
© 2009 G. W. Roberts Op Amp Circuits, slide 121
304-330 Introduction To Electronic Circuits
Max. Output Voltage Level, Gain and BW Trade-Off
BW0 0 ω tω t10
Gai
n
1
10
100
ω t100
Gain ⋅BW =ω t
Max
. Out
put V
olta
ge
Leve
l
V̂1
V̂100
V̂10
V̂ ⋅BW = SR
© 2009 G. W. Roberts Op Amp Circuits, slide 122
304-330 Introduction To Electronic Circuits
Maximum Input Condition
BW0 0 ω tω t10
ω t100
Gain ⋅BW =ω t
Max
. Inp
ut V
olta
ge
Leve
l
VIN ,max
V̂ ⋅BW = SRVIN ,max =
SRω t
When combined, one gets
Independent of frequency!
© 2009 G. W. Roberts Op Amp Circuits, slide 123
304-330 Introduction To Electronic Circuits
Assignment #2 (partial) • Sedra/Smith:
– 2.5, 2.6, 2.8, 2.9, 2.12, 2.15, 2.20, 2.34, 2.41,– 2.44, 2.50, 2.51, 2.53, 2.60, 2.72, 2.74, – 2.79, 2.80, 2.85, 2.89,– 2.98, 2.104, 2.106,– 2.107, 2.109, 2.112, 2.115, 2.119, 2.126
© 2009 G. W. Roberts Op Amp Circuits, slide 124
304-330 Introduction To Electronic Circuits
Outline• The Ideal Op Amp• The Inverting Configuration• The Noninverting Configuration• Difference Amplifiers• Integrators and Differentiators• DC Imperfections• Effects of Finite Open-Loop Gain and Bandwidth
on Circuit Performance• Large-Signal Operation of Op Amps• Simulating Op Amp Circuits With SPICE• Summary
© 2009 G. W. Roberts Op Amp Circuits, slide 125
304-330 Introduction To Electronic Circuits
Op Amp Model(VCVS Model)
• A SPICE simulation of any op amp circuit begins by replacing the op amp by the simple VCVC equivalent representation.
A(s) vo
+
- =
Aovi
+
- ± vo
+
- vi vi
+
-
© 2009 G. W. Roberts Op Amp Circuits, slide 126
304-330 Introduction To Electronic Circuits
Inverting Amplifier Example
• Here a transfer function (.TF) analysis is requested.– A .TF command determines the 2-port representation of
the circuit at DC.
Op Amp Circuit
SPICE Input Deck
Inverting Amplifier Configuration
** Circuit Description **
* signal sourceVi 3 0 DC 1v* inverting amplifier circuit descriptionR1 3 2 1kR2 2 1 10kEopamp 1 0 0 2 1e6
** Analysis Requests **.TF V(1) Vi
** Output Requests *** none required
.end
- +
1 kΩ
€
vO± vI
12
3
10 kΩ
© 2009 G. W. Roberts Op Amp Circuits, slide 127
304-330 Introduction To Electronic Circuits
Inverting Amplifier Example
.TF Analysis Output
vI 1 kΩ+
- ± vo +
- -10vi
Op Amp Circuit
- +
1 kΩ
€
vO± vI
12
3
10 kΩ
© 2009 G. W. Roberts Op Amp Circuits, slide 128
304-330 Introduction To Electronic Circuits
Subcircuits In SPICE
• Subcircuits are used in SPICE to simplify coding.• Subcircuits act as subroutines in SPICE.
Calling Statement In Main Routine
Subcircuit Description
Inverting Amplifier Configuration
** Circuit Description **
* signal sourceVi 3 0 DC 1v* inverting amplifier circuit descriptionR1 3 2 1kR2 2 1 10kXop 1 0 2 opamp
** Analysis Requests **.TF V(1) Vi
** Output Requests *** none required
.end
* op-amp subcircuit.subckt opamp 1 2 3* connections: | | |* output | |* +ve input |• -ve input
Eoutput 1 0 2 3 1e6
.ends opamp
© 2009 G. W. Roberts Op Amp Circuits, slide 129
304-330 Introduction To Electronic Circuits
Op Amp Model(VCVS Model With Input and Output Loading Effects)
• To account for op amp loading effects, we can add input and output resistances as shown above.
• Input offset voltage effects can also be included in the model.
A(s) vo
+
- =
Aovi
+
- ± vo
+
- vi vi
+
- Rin
Rout - +
Voffset
© 2009 G. W. Roberts Op Amp Circuits, slide 130
304-330 Introduction To Electronic Circuits
Op Amp Model(Finite Bandwidth Model)
A(s) vo
+
- = v1
R
C Aovi
+
-
+
- ±
ω0
1
AO
€
Vo
Vi
€
ωb =1RC
€
ω t =AORC
€
A s( ) =Vo
Vi
=AO
1+ sωb
=AO
1+ sRC
€
A jω( ) ≈ AOωRC
for ω >>ωb
± vo +
- vi v1
vi
+
-
© 2009 G. W. Roberts Op Amp Circuits, slide 131
304-330 Introduction To Electronic Circuits
Subcircuit For BW Limited Op Amp
Subcircuit
Subcircuit Description
* op-amp subcircuit.subckt opamp 1 2 3* connections: | | |* output | |* +ve input |• -ve input
E1 4 0 2 3 1e6E2 1 0 5 0 1R 4 5 1e3C 5 0 1e-6
.ends opamp
12
3
4 5
v1
1 kΩ
1 mF 106 vi
+
-
+
- ± ±
vo +
- vi v1
© 2009 G. W. Roberts Op Amp Circuits, slide 132
304-330 Introduction To Electronic Circuits
Op Amp Model(Finite Bandwidth & Output Limited)
A(s) vo
+
- =
v1
R
C Aovi
+
-
+
- ± ±
vo +
- vi v1
vi
+
-
± ±
1 Ω
vo,max
vi
vo,max vO
0,0
Ao
vo,max
-vo,max
−vo,maxAo
vo,maxAo
Ideal Diode (more next chapter)
© 2009 G. W. Roberts Op Amp Circuits, slide 133
304-330 Introduction To Electronic Circuits
Op Amp Model(Slew-Rate / BW Model)
A(s) vo
+
- = vi
+
-
v2
R
C Aov1
+
-
+
-± ±
vo
+
-vi
v2
± ±
1 Ω
vo,max
vi
vO
0,0
Ao
±v1
+
-vi
± ±
1 Ω
SR = Ao ⋅1RC
⋅ v1,max ⇒ v1,max = SR× R×CAo
=SR
Ao ×ωb
=SRωt
vo,max
vo,max
-vo,max
v1,max v1,max
−v1,maxv1,max
© 2009 G. W. Roberts Op Amp Circuits, slide 134
304-330 Introduction To Electronic Circuits
Assignment #2 (complete)• Sedra/Smith:
– 2.5, 2.6, 2.8, 2.9, 2.12, 2.15, 2.20, 2.34, 2.41,– 2.44, 2.50, 2.51, 2.53, 2.60, 2.72, 2.74, – 2.79, 2.80, 2.85, 2.89,– 2.98, 2.104, 2.106,– 2.107, 2.109, 2.112, 2.115, 2.119, 2.126
• Roberts/Sedra: – 2.5, 2.6, 2.7, 2.8, 2.9
© 2009 G. W. Roberts Op Amp Circuits, slide 135
304-330 Introduction To Electronic Circuits
Summary• The op amp is a versatile circuit building block. It is
easy to apply and the performance of op amp circuits closely matches theoretical predictions.
• The ideal op amp responds only to the difference input signal (v2-v1) and provides at the output w.r.t. ground a signal A(v2-v1), where A is the open loop gain.
• Op amps circuit work because of negative feedback.